The question is in the title : Can we find spaces $A$ and $B$, each non contractible, such that their smash product $A \wedge B$, i.e. the homotopy cofibre of $A \vee B \to A \times B$, is a contractible space ?
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4$\begingroup$ Maybe. Take $A=S^1$ and $B$ the classifying space of a group with trivial homology, but I don't know if these exist. $\endgroup$– Fernando MuroCommented Sep 23, 2015 at 15:14
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4$\begingroup$ @FernandoMuro: there are many groups with trivial homology. There is a general construction as part of the proof of the Kan-Thurston theorem, for example. $\endgroup$– Neil StricklandCommented Sep 23, 2015 at 15:31
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2 Answers
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Oh yes ! Fernando gave the hint ! Take $A = S^1$ and $B$ = the Epstein's space. Then $A \wedge B \simeq \Sigma B$ is contractible but $B$ is not !
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$\begingroup$ What's the Epstein's space? $\endgroup$ Commented Sep 26, 2015 at 18:42
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$\begingroup$ Here is the reference for Epstein's space : D.B.A.Epstein, "A group with zero homology", Mathematical Proceedings of the Cambridge Philosophical Society 64(03):599 - 602 · July 1968 $\endgroup$– jpaulCommented Sep 30, 2015 at 10:31
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$\begingroup$ Thank you. From your comment, I understand that Epstein's space is the classifying space for a certain group. $\endgroup$ Commented Sep 30, 2015 at 20:47
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I just saw this in Ravenel's "orange book"! Let $X$ be any simply connected CW complex whose reduced homology is all torsion. Let $X_{(p)}$ be the p-localization for a prime $p$. Then for primes $p\neq q$, $X_{(p)}\wedge X_{(q)}$ is contractible. But neither $X_{(p)}$ nor $X_{(q)}$ is contractible if $H_*(X)$ has $p$ and $q$ torsion.
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6$\begingroup$ In the same spirit, just take the smash product of two CW complexes that are Moore spaces for finite cyclic groups of relatively prime orders. $\endgroup$ Commented Sep 23, 2015 at 23:50
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$\begingroup$ Oh yeah, that's much simpler :-) $\endgroup$ Commented Sep 24, 2015 at 2:08